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LenKa [72]
3 years ago
6

Suppose that you draw 3 cards without replacement from a 52 card deck.

Mathematics
2 answers:
Alenkinab [10]3 years ago
5 0

Answer:

0.000181

Step-by-step explanation:

Andru [333]3 years ago
3 0

Answer:

0.000181

Step-by-step explanation:

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What is the answers to the picture
nalin [4]

5.6/4 = 1.4(4)/4

x = 4

4 0
3 years ago
Read 2 more answers
Marge cut 16 pieces of tape for mounting pictures on posterboard each piece of tape is 3/8 inches long how much tape did Marge u
daser333 [38]

Answer:

6 inches of tape

Step-by-step explanation:

To find the total amount of tape that Marge used, you can take the measure of each piece (\frac{3}{8}) and multiply by the total number of pieces cut (16):

\frac{3}{8}*\frac{16}{1}=\frac{48}{8}=6inches

So, Marge used 6 inches of tape altogether.

7 0
2 years ago
I really need help with this it’s on Aleks
liraira [26]

Answer:

y = 2

Step-by-step explanation:

Given

\sqrt[3]{3y+2} + 4 = 6 ( subtract 4 from both sides )

\sqrt[3]{3y+2} = 2 ( cube both sides )

3y + 2 = 2³ = 8 ( subtract 2 from both sides )

3y = 6 ( divide both sides by 3 )

y = 2

6 0
3 years ago
Read 2 more answers
Find the derivative.
Aleksandr [31]

Answer:

Using either method, we obtain:  t^\frac{3}{8}

Step-by-step explanation:

a) By evaluating the integral:

 \frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du

The integral itself can be evaluated by writing the root and exponent of the variable u as:   \sqrt[8]{u^3} =u^{\frac{3}{8}

Then, an antiderivative of this is: \frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}

which evaluated between the limits of integration gives:

\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}

and now the derivative of this expression with respect to "t" is:

\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

g(x)=\int\limits^x_a {f(t)} \, dt

is continuous on [a,b], differentiable on (a,b) and  g'(x)=f(x)

Since this this function u^{\frac{3}{8} is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}

5 0
3 years ago
GUYS I REALLY NEED HELP I DON'T UNDERSTAND :"(
ANTONII [103]

Answer:

the answer is 7

Step-by-step explanation:

plot the points then count how many spaces there are going diagnal

4 0
3 years ago
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